The leading coefficients of the degree-$n$ elements of an ideal of $R[x]$, together with $0$, form an ideal of $R$, and these ideals ascend with $n$

Fix $n\in \mathbb{N}$ and read the displayed definition: ${\mathfrak{a}}_n$ consists of $0$ together with the leading coefficients of those elements of $\mathfrak{a}$ that are nonzero of degree exactly $n$. In particular $0\in {\mathfrak{a}}_n$, so ${\mathfrak{a}}_n$ is a nonempty subset of $R$; and no element of ${\mathfrak{a}}_n$ other than the adjoined $0$ is $0$, since a leading coefficient is nonzero by definition.

${\mathfrak{a}}_n$ is closed under differences. Let $c,c^{\mathrm{\prime }}\in {\mathfrak{a}}_n$. If $c^{\mathrm{\prime }}=0$ then $c-c^{\mathrm{\prime }}=c\in {\mathfrak{a}}_n$. If $c=0$ and $c^{\mathrm{\prime }}\mathrm{\ne }0$, take $g\in \mathfrak{a}$ realising $c^{\mathrm{\prime }}$ at stage $n$; negation is coefficientwise, so $-g\in \mathfrak{a}$ is nonzero of degree $n$ with $\mathrm{lc}(-g)=-c^{\mathrm{\prime }}$, and $c-c^{\mathrm{\prime }}=-c^{\mathrm{\prime }}\in {\mathfrak{a}}_n$. If both are nonzero, take realisers $f,g\in \mathfrak{a}$ at stage $n$; then $-g$ is nonzero of degree $n$, the coefficient of $x^n$ in $f-g$ is $c-c^{\mathrm{\prime }}$, and the coefficients of $f-g$ above index $n$ all vanish. Should $c-c^{\mathrm{\prime }}=0$, the difference lies in ${\mathfrak{a}}_n$ as the adjoined element; otherwise $f-g\mathrm{\ne }0$, the degree law gives $\mathrm{deg}(f-g)\le \max \{n,n\}=n$, and the nonvanishing coefficient at $x^n$ forces $\mathrm{deg}(f-g)=n$ with $\mathrm{lc}(f-g)=c-c^{\mathrm{\prime }}$, so $c-c^{\mathrm{\prime }}\in {\mathfrak{a}}_n$.

${\mathfrak{a}}_n$ is closed under multiplication by elements of $R$. Let $c\in {\mathfrak{a}}_n$ and $r\in R$. If $rc=0$ the product lies in ${\mathfrak{a}}_n$ as the adjoined element, and this covers $c=0$ and $r=0$. Otherwise $c\mathrm{\ne }0$ and $r\mathrm{\ne }0$; take $f\in \mathfrak{a}$ realising $c$ at stage $n$ and read $r$ as a constant polynomial, which is nonzero of degree $0$ with leading coefficient $r$. The coefficient of $x^{0+n}$ in the product $rf$ is $rc\mathrm{\ne }0$, so $rf\mathrm{\ne }0$; the degree law then gives $\mathrm{deg}(rf)\le 0+n$, and the nonvanishing coefficient at $x^n$ forces $\mathrm{deg}(rf)=n$ with $\mathrm{lc}(rf)=rc$. Since $\mathfrak{a}$ is an ideal of $R[x]$ we have $rf\in \mathfrak{a}$, so $rc\in {\mathfrak{a}}_n$.

The stages ascend. Let $c\in {\mathfrak{a}}_n$ with $c\mathrm{\ne }0$ and let $f\in \mathfrak{a}$ realise it at stage $n$. Multiplying by $x$ shifts coefficients: by the convolution rule the coefficient of $x^i$ in $xf$ is the coefficient of $x^{i-1}$ in $f$ for $i\ge 1$ and is $0$ at $i=0$. So $xf$ has zero coefficients above index $n+1$ and coefficient $c\mathrm{\ne }0$ at index $n+1$, whence $xf$ is a nonzero element of $\mathfrak{a}$ of degree $n+1$ with $\mathrm{lc}(xf)=c$. Thus $c\in {\mathfrak{a}}_{n+1}$, and $0\in {\mathfrak{a}}_{n+1}$ as well, so ${\mathfrak{a}}_n\subseteq {\mathfrak{a}}_{n+1}$.

By the ideal criterion, a nonempty subset of a commutative ring closed under differences and under multiplication by arbitrary ring elements is an ideal; steps 2.1 and 2.2 supply exactly those closures, so ${\mathfrak{a}}_n$ is an ideal of $R$.

Step 3.1 holds for every $n\in \mathbb{N}$ and step 2.3 gives ${\mathfrak{a}}_n\subseteq {\mathfrak{a}}_{n+1}$ for every $n\in \mathbb{N}$, so the stages form an ascending chain of ideals of $R$ indexed by $\mathbb{N}$ and beginning at ${\mathfrak{a}}_0$. A nonzero element of $\mathfrak{a}$ has degree $0$ exactly when it is a nonzero constant, and its leading coefficient is then that constant, so ${\mathfrak{a}}_0$ is the set of constants lying in $\mathfrak{a}$, the zero constant included.